For the reaction shown, calculate the theoretical yield of the product in grams for each of the initial quantities of reactants. $$ \mathrm{Ti}(s)+2 \mathrm{~F}_{2}(g) \longrightarrow \mathrm{TiF}_{4}(s) $$ (a) \(1.0 \mathrm{~g} \mathrm{Ti} ; 1.0 \mathrm{~g} \mathrm{~F}_{2}\) (b) \(4.8 \mathrm{~g} \mathrm{Ti} ; 3.2 \mathrm{~g} \mathrm{~F}_{2}\) (c) \(0.388 \mathrm{~g} \mathrm{Ti} ; 0.341 \mathrm{~g} \mathrm{~F}_{2}\)

When it comes to chemical reactions, stoichiometry is the mathematical relationship between the quantities of reactants and products. Think of it as a recipe in chemistry, where we need just the right amount of each reactant to make the desired product. By knowing the balanced chemical equation, we can figure out how different amounts of reactants relate to the amount of products formed.

In our textbook exercise, stoichiometry allows us to calculate how much product, Titanium tetrafluoride (TiF₄), can be made from given amounts of Titanium (Ti) and Fluorine gas (F₂). This is essential in predicting the output of a reaction and in ensuring that we utilize our reactants efficiently, which is vital in both laboratory and industrial settings.

A chemical reaction can only proceed until the limiting reactant is completely used up, making it the ingredient that determines the extent of the reaction. This concept is similar to running out of flour while baking cookies; no matter how much of the other ingredients you have, you can't make more cookies without it.

In the exercise, determining the limiting reactant between Titanium and Fluorine gas is crucial for calculating the theoretical yield of Titanium tetrafluoride. We compare the mole ratio of reactants provided to the mole ratio required by the balanced chemical equation. The reactant that has fewer moles than required, according to this stoichiometric ratio, is the limiting reactant. This step ensures precise and accurate calculations of the theoretical yield in chemistry problems.

The molar mass is the weight of one mole of a substance, usually expressed in grams per mole (g/mol), and it's a bridge between the atomic scale and the human scale. For atoms, the molar mass is determined using the atomic weight from the periodic table, while for molecules, it's the sum of the atomic weights of each atom in the molecule.

Understanding the molar mass is essential, as shown in the solution steps of our exercise. It is the key to converting between moles and grams, which allows us to measure out the correct amounts of each reactant and to determine the theoretical yield of the product. For example, knowing the molar mass of Titanium tetrafluoride enables us to convert the moles of TiF₄ we expect to make to the actual mass of product that could be formed from the reaction.